Driving SPDT relay using PIC18F2580

[Gaara]

Hi all,

I’m working on driving a relay i.e. the Omron G5Q-1A with a PIC18F2580 using a Darlington transistor array specifically the ULN2803A. I’m utilising the ULN2803A as I need to drive more than one relay but I’m starting with the driving of a single relay. For testing I’m connecting an LED which I intend on toggling using the PIC as my relay load. My basic objective is to use this relay to toggle an LED connected to the Normally Open pin of the relay.

I’ll describe the hardware side of things before moving into the code.

Firstly the details of my relay are as follows:
Model: Omron G5Q-1A
Coil resistance: 63ohms.
Contact form: SPDT- 1C

Here is my schematic: https://1d563b1a-a-62cb3a1a-s-sites.googlegroups.com/site/bgedsadownloads/Relay%20driving.png?attachauth=ANoY7coZglmInjceNzdOGPcfj-1dzKHq0ZqorQcDte_pGTZPnO7NOWRugChZb0tyxTswt_JUV6-LYmcoxSeQcmNvgzWSpwjo_avqv04e6gs9jPkfpSfuvPnbskvrSX3bTU1QjgqcXmXi-2VHY1rpy_YtHNzmsSixAeXiLMgIT2YOh0t2chNGYhm22aOyzWsh0gcTxt0iAeIN2tmo3t6u42ekn8e9yYLs0Q%3D%3D&attredirects=0

With regards to the relay I can confirm I’ve connected the coil pins i.e. pins 1 and 5 of the relay as per page 3 of the relay datasheet (http://www.omron.com/ecb/products/pdf/en-g5q.pdf
) correctly. I tested these two pins for resistance as they are the only pins which have resistance as of when the relay is not connected. I obtained a resistance of 60ohms which was very closed to the 63ohms for SPDT (Ic) relay as per specified in the relay datasheet.

Also using the diagram on page 3 of the relay datasheet I can confirm that the pole pin is pin 4. As seen in my schematic I then connected the LED on my NC pin which should be engaged when the relay switches using its pole.

Before posting this up I read through the following sources regarding the setup i.e.:
theory behind the darlington transistor array: http://www.electronics-tutorials.ws/transistor/tran_2.html
connection between a PIC and a darlington transistor array: http://electronics.stackexchange.com/questions/36629/bjt-resistor-and-diode-work-but-uln2803-does-not
SPDT based relays: http://www.zen22142.zen.co.uk/ronj/cpr.html

My relay’s datasheet: http://www.omron.com/ecb/products/pdf/en-g5q.pdf
ULN2803A datasheet: http://www.ti.com/lit/ds/symlink/uln2803a.pdf

Regarding the software aspects below is my code that I intend on using to achieve the basic objective I mentioned earlier i.e. is to use this relay to toggle an LED connected to the Normally Open pin of the relay. If you notice from my schematic I’ve got an opto-isolated LED connected to the PIC. This LED is being used to indicate that the respective relay pin has been pulled high or low. The code below toggles the LED connected to PIN C1 of the PIC.

[Ttelmah]

Three things one after the other:

1) If a pin won't do what you expect, look what else is on it. In your case, RC1, is the timer1 oscillator input. If you look in the data sheet at the I/O summary for PortC, you see that this 'overrides the TRIS control when enabled'. So start by disabling Timer1:

setup_timer_1(T1_DISABLED);

May solve the I/O problem.

2) Are you sure about your part number?!..... The ULN2803, is an octal Darlington driver as you describe. The 2308, is an I2C interface driver.... Your PNG, has 2308.

3) Now key thing to understand when switching anything inductive, is that when you switch 'off', there is significant magnetic energy in the coil. This has to go somewhere. The diodes built into the chip, dump this into the supply rail. There has to be sufficient load/capacitance on this rail to prevent it from rising significantly when this happens. You only show 0.1uF on this rail.....

[Gaara]

Hi Ttelmah,

Thanks for the reply.

[alan]

According to your schematic you have 2 LED's on RB2 and RB3. Use them to do a 1Hz test to ensure your PIC are running at the correct speed.

Regards

[Gaara]

Hi alan

I believe the PIC is running at the correct speed as the LED I call REL_1 in my code (i.e. the relay connected to the optocoupler which is connected to PIN_C1 ) is toggling at the correct speed.

Thanks
Gaara

[ezflyr]

Hi,

Good lord, I never imagined that such a simple topic could be made so complicated......

You really need to work on your basic troubleshooting skills!

First, what is the coil voltage of your relay? Are you sure? OK, with your relay out of the circuit, use a power supply of the correct value to activate the coil. You should hear a faint 'click' when you do this. Does that work? If not, reverse the polarity of the applied voltage. Is it working now? If so, put a continuity tester across the NO terminals and activate the coil again with your power supply. Does your continuity tester beep?

Once you verify the proper operation of the relay, and verify you are connecting to the correct pins, now try to drive the relay with the ULN3803 driver chip. At this point, just control the input to the driver manually, ie. simulate the output of your PIC by connecting it to GND and +5V. Does this work?

The point of my post is that you have all this stuff right in front of you, so YOU are best able to figure out what is wrong by breaking the problem down and solving it step-by-step. You are either going to fix the problem, or find out what is the issue......

John

PS Would you ask your 'alter-ego', TokTok, if he managed to solve his communications problem with the Zigbee module?

[Mike Walne]

Mike

A further thought.

You can check the relay operates (in circuit with the power on) by simply connecting from the correct LM2803 output to GND.

Incidentally the protection diode does not return energy to the supply rail.
When you turn the relay off the coil current simply circulates through the coil and the diode.
The energy is dissipated in the diode and the coil's resistance.
The time taken depends on the initial current, the coil's resistance & inductance and the diode's forward drop.
To return energy to the supply, via the diode, you would need to connect the other end of the coil to GND.

Like ezflyr says you should be able to do these simple tests on your own.

Mike

[Ttelmah]

Apologies for the phrasing. They return energy _through_ the connections between the drivers diode pin and the drivers output. This results in spikes between these dependant on how good the connection is. They also require that the supply can deliver the current change when they turn on/off, without drooping.

The poster doesn't show his supply.

0.1uF, is a 'good' capacitor to have nice and close to the PIC (don't get rid of this), but the main supply needs to be able to have a 'step change' of 200mA, without changing by more than perhaps 0.1v. Depends on the impedance and speed of response of the supply, but several hundred uF, would be typically needed.

Just how large the spikes can be, and the effect on circuitry, can be surprising.

I prefer to have the trap pin connected independently of the processor supply. Using a logic supply to switch relays can lead to a lot of problems...

[Mike Walne]

I know this is NOT a circuit analysis forum but here's a take on relay switching.
With the switch closed the coil current ~80mA flows as shown. (Using Gaara's 63Ohm relay coil. )
From supply +ve, via lead inductance, coil inductance & resistance, switch, return lead inductance, back to supply -ve

[Ttelmah]

Which is why I prefer to separate logic and relay supplies.

Remember this can (to a limited extent), be as simple as having one 5v rail, feeding the relays, and then having a resistor/capacitor filter feeding the logic off this.

And (of course), concentrate on how the ground rails run. You don't want the relay current being drawn 'via' the logic ground....

[Gaara]

Hi all,

@ezflyr:

thanks for the reply.

[Mike Walne]

This is getting silly, and very tedious.
You claim you're using Omron G5Q-1A relay.
You insist it's SPDT
You've provided link to Omron data sheet which says SPST.

[Ttelmah]

Also, no mention of the rating of the '6v PSU', or even if this is regulated, or not....

[gpsmikey]

And the problem with the 6v supply regulated to 5v with that regulator is if you look at the spec. from the sheet posted, the dropout may be up to 1.3 volts ... that means the input to the regulator MUST be at LEAST 5.0 + 1.3 = 6.3v. And that assumes NO noise or ripple etc on the input side. At 6.0v on the input, you have typically already dropped out of regulation and are asking for nothing but trouble.

mikey
_________________
mikey
-- you can't have too many gadgets or too much disk space !
old engineering saying: 1+1 = 3 for sufficiently large values of 1 or small values of 3

[avro698]

Referring to your posted schematic, what's connected to the wiper (pin 2) of the relay? All I see is a label, X17-2. If this is floating, the LED will never toggle.
If this is the case, consider connecting it to GND and the LED should toggle (assuming the relay energizes / denergizes).

FF101